Question

question 11 · 1 point evaluate each of the following limits, then identify any vertical asymptotes of the function (f(x)=\frac{1}{(x - 4)^5}). if any of the limits do not exist, enter (varnothing) as your answer. (lim_{x
ightarrow4^{-}}\frac{1}{(x - 4)^5}) (lim_{x
ightarrow4^{+}}\frac{1}{(x - 4)^5}) (lim_{x
ightarrow4}\frac{1}{(x - 4)^5}) a. (lim_{x
ightarrow4^{-}}f(x)=)

Explanation:

Step1: Analyze left - hand limit

We want to find $\lim_{x
ightarrow4^{-}}\frac{1}{(x - 4)^5}$. As $x
ightarrow4^{-}$, then $x-4
ightarrow0^{-}$. Let $t=x - 4$, so when $x
ightarrow4^{-}$, $t
ightarrow0^{-}$. Then $\frac{1}{(x - 4)^5}=\frac{1}{t^5}$. Since $t
ightarrow0^{-}$, $t^5
ightarrow0^{-}$ (because an odd - power of a negative number is negative), and $\frac{1}{t^5}
ightarrow-\infty$.

Step2: Analyze right - hand limit

We want to find $\lim_{x
ightarrow4^{+}}\frac{1}{(x - 4)^5}$. Let $t=x - 4$, when $x
ightarrow4^{+}$, $t
ightarrow0^{+}$. Then $\frac{1}{(x - 4)^5}=\frac{1}{t^5}$. Since $t
ightarrow0^{+}$, $t^5
ightarrow0^{+}$ (because an odd - power of a positive number is positive), and $\frac{1}{t^5}
ightarrow+\infty$.

Step3: Analyze two - sided limit

Since $\lim_{x
ightarrow4^{-}}\frac{1}{(x - 4)^5}=-\infty$ and $\lim_{x
ightarrow4^{+}}\frac{1}{(x - 4)^5}=+\infty$, $\lim_{x
ightarrow4}\frac{1}{(x - 4)^5}$ does not exist.

Step4: Find vertical asymptote

A vertical asymptote occurs at $x = a$ if $\lim_{x
ightarrow a^{-}}f(x)=\pm\infty$ or $\lim_{x
ightarrow a^{+}}f(x)=\pm\infty$. For the function $f(x)=\frac{1}{(x - 4)^5}$, since $\lim_{x
ightarrow4^{-}}f(x)=-\infty$ and $\lim_{x
ightarrow4^{+}}f(x)=+\infty$, the vertical asymptote is $x = 4$.

Answer:

a. $\lim_{x
ightarrow4^{-}}f(x)=-\infty$
$\lim_{x
ightarrow4^{+}}f(x)=+\infty$
$\lim_{x
ightarrow4}f(x)=\varnothing$
Vertical asymptote: $x = 4$